(1)函数f(x)的值域为(-1,+∞),由y=2x-1,得 x=log2(y+1), 所以f-1(x)=log2(x+1)(x>-1),任取-1<x1<x2, f-1(x1)-f-1(x2)=log2(x1+1)-log2(x2+1)=log2, 由-1<x1<x2得0<x1+1<x2+1,因此0<<1,得 log2<0, 所以f-1(x1)<f-1(x2),故f-1(x)在(-1,+∞)上为单调增函数. (2)f-1(x)≤g(x) 即:log2(x+1)≤log4(3x+1)⇔⇔, 解之得0≤x≤1,所以D=[0,1]. (3)H(x)=g(x)-f-1(x)=log4(3x+1)-log2(x+1)=log2=log2(3-), 由0≤x≤1,得1≤3-≤2,所以0≤log2(3-)≤,因此函数H(x)的值域为[0,]. |